Question

Suppose that scores on a test are normally distributed with a mean of 80 and a standard deviation of 8. Answer the questions below. (a) What is the 70th percentile? (round to the tenths place) (b) What percentage of students score less than 70? (round to the tenths place, give the percent)

Answer 1

Answer:

(a) 84.2

(b) 10.6

Step-by-step explanation:

To solve this questions we can use the standardization formula, where we know that if $X\sim N(\mu,\sigma^2)$ then $Z=\frac{X-\mu}{\sigma} \sim N(0,1)$

So for

(a) we know that the z score for the 70th percentile is 0.524, so using the normalization equation we have

$\frac{X-\mu}{\sigma}=0.524$

$X=0.524*8+80=84.192$

(b) We can procede as above and get

$P(X$

Answer 2

Answer:

a) 84.2  b)10.6%

Step-by-step explanation:

Hi there! Hope you are doing fine.

First to obtain the 70th percentile we make use of tables found (everywhere) with the relation bewteen the percentile and the so-called z-score

Such like this one:

https://www.mymathtables.com/statistic/z-score-percentile-normal-distribution.html

There we look for the 70th percentile and found that its z-score is 0.524. This is the number of times of standar deviations which we are far from the mean, i.e. :

$z = \frac{x- \mu}{\sigma}$ -- (1)

In our case we have:

μ = 80

σ = 8

replacing these values on the eq (1) we have:

$0.524 = \frac{x- 80}{8}$

Now we find x

$x = 0.524*8+80$

So x=84.192 and if we round it to the tenths place:

     x=84.2

For the second part we now look for the percentile of a z-score with x = 70

now:

$z = \frac{70-80}{8} = -1.25$

Now we look in the table for a number close to it or alternatively we can use a calculator such as this one:

https://measuringu.com/pcalcz/

   Just make sure you select the one-sided option

We found that the -1.25 z-score corresponds to 10.565%

Rounding it to the tenths position we get

   p = 10.6%